# Calculate the Energy that Can Be Obtained from 1 Kg of Water Through the Fusion Reaction 2h + 2h → 3h + P. Assume that 1.5 × 10−2% of Natural Water is Heavy Water D2o (By Number of Molecules) - Physics

Sum

Calculate the energy that can be obtained from 1 kg of water through the fusion reaction 2H + 2H → 3H + p. Assume that 1.5 × 10−2% of natural water is heavy water D2O (by number of molecules) and all the deuterium is used for fusion.

(Use Mass of proton mp = 1.007276 u, Mass of ""_1^1"H" atom = 1.007825 u, Mass of neutron mn = 1.008665 u, Mass of electron = 0.0005486 u ≈ 511 keV/c2,1 u = 931 MeV/c2.)

#### Solution

Given :-

18 g of water contains 6.023 xx 10^23 molecules.

∴ "1000 g of water" = (6.023 xx 10^23 xx 1000)/18 = 3.346 xx 10^25 molecules

% "of deuterium" = 3.346 xx 10^25 xx 0.015/100 = 0.05019 xx 10^23

Energy of deuterium = 30.4486 xx 10^25

= [2 xx m(""^2H) - m(""^3H) - m_p]c^2

= (2 xx 2.014102  "u" - 3.016049  "u" - 1.007276   "u")c^2

= 0.004879 xx 931 "MeV"

= 4.542349" MeV"

= 7.262 xx 10^-13 "J"

Total energy = 0.05019 xx 10^23 xx 7.262 xx 10^-13  "J"

= 3644 MJ

Concept: Nuclear Energy - Fission
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 24 The Nucleus
Q 53 | Page 444